$$\begin{aligned} x^{\prime} &= \gamma \left( x-vt \right) \\ y^{\prime} &= y \\ z^{\prime} &= z \\ t^{\prime} &= \gamma \left( t-\frac{vx}{c^{2}} \right) \end{aligned}$$

These equations correspond to the relatively simple case in which the relative motion of the two observers is along their common x-axes. Of course, other directions of motion are possible, but the most general Lorentz transformation is rather complicated, with all four quantities mixed up together. We shall continue to use this simpler form, since it contains all the essential features of relativity.

Let us now discuss more of the consequences of this transformation. First, it is interesting to solve these equations in reverse. That is, here is a set of linear equations, four equations with four unknowns, and they can be solved in reverse, for x, y, z, t in terms of x’, y’, z’, t’. The result is very interesting, since it tells us how a system of coordinates “at rest” looks from the point of view of one that is “moving”. Of course, since the motions are relative and of uniform velocity, the man who is “moving” can say, if he wishes, that it is really the other fellow who is moving and he himself who is at rest. And since he is moving in the opposite direction, he should get the same transformation, but with the opposite signs of velocity. This is precisely what we find by manipulation, so that is consistent.

Steps: To make x the subject, unprimed the prime and prime the unprime and add negative to v.

Relativistic equations should reduce to familiar classical equations if we let c approach infinity. If we let c → ∞, γ → 1.

$$\begin{aligned} x &= \gamma \left( x^{\prime} + v t^{\prime} \right) \\ t &= \gamma \left( t^{\prime} + \frac{vx^{\prime}}{c^{2}} \right) \end{aligned}$$

To find the differences between coordinates for a pair of events: